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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/chemistry/ocr/a/haloalkanes/flashcards","label":"Haloalkanes","subjectCode":"a_level_chemistry__ocr_a","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"d47bc91c-fe3d-47cf-9baf-a5e0a6cc40aa","title":"Haloalkanes Flashcards","topicId":"45c20e17-9977-57ba-8d0e-b91405e3f738","topicSlug":"haloalkanes","cards":[{"id":"e852dfe6-2967-438f-b37a-f4223d5e61c8","cardType":"basic","points":1,"front":"What is a nucleophile?","back":"An electron pair donor.","mcqOptions":[{"text":"An electron pair donor.","correct":true,"explanation":"A nucleophile donates a pair of electrons to an electron-deficient atom, forming a new covalent bond. The donated pair may come from a lone pair or a negative charge."},{"text":"An electron pair acceptor.","correct":false,"explanation":"An electron pair acceptor is an electrophile, not a nucleophile. Nucleophiles are electron-rich species that donate an electron pair to form a covalent bond."},{"text":"A proton donor.","correct":false,"explanation":"A proton donor is a Brønsted-Lowry acid, whereas a nucleophile donates an electron pair to an electron-deficient atom during a reaction."},{"text":"A proton acceptor.","correct":false,"explanation":"A proton acceptor is a Brønsted-Lowry base, but the defining feature of a nucleophile is donation of an electron pair to form a bond."}]},{"id":"76da963c-9af9-4db5-a502-8f4ba24cf076","cardType":"cloze","points":null,"front":"For an atom to act as a nucleophile, it must possess at least one [ ].","back":"For an atom to act as a nucleophile, it must possess at least one lone pair of electrons.","mcqOptions":[{"text":"lone pair of electrons","correct":true,"explanation":"A lone pair of electrons is a non-bonding pair available for donation to an electron-deficient atom, allowing the atom to act as a nucleophile."},{"text":"bonding pair of electrons","correct":false,"explanation":"A bonding pair of electrons is shared between two atoms and is not normally available for nucleophilic donation. The required feature is at least one lone pair of electrons."},{"text":"partial positive charge","correct":false,"explanation":"A partially positive charge attracts electron-rich species but does not provide electrons for donation. Nucleophiles require at least one lone pair of electrons."},{"text":"empty orbital","correct":false,"explanation":"An empty orbital can accept an electron pair and is characteristic of an electrophile. A nucleophile instead must possess at least one lone pair of electrons."}]},{"id":"cbb0b29e-6deb-4b0a-af88-5b2da5fe8c2d","cardType":"basic","points":3,"front":"Explain the polarity of the $C-X$ bond in a haloalkane.","back":"

Halogen is more electronegative than carbon
Electron density is pulled toward the halogen
Carbon becomes $\\delta+$ (electron-deficient) and halogen becomes $\\delta-$

","mcqOptions":[{"text":"

Halogen is more electronegative than carbon
Electron density is pulled toward the halogen
Carbon becomes $\\delta+$ (electron-deficient) and halogen becomes $\\delta-$

","correct":true,"explanation":"The halogen is more electronegative than carbon, so it attracts the bonding electron density. This gives carbon a δ+ charge and the halogen a δ- charge."},{"text":"

Carbon is more electronegative than halogen
Electron density is pulled toward the carbon
Carbon becomes $\\delta-$ (electron-rich) and halogen becomes $\\delta+$

","correct":false,"explanation":"Carbon is less electronegative than the halogen, so electron density is not pulled towards carbon. The correct polarity has carbon δ+ and the halogen δ-."},{"text":"

Halogen is equally electronegative as carbon
Electron density is shared evenly between the atoms
Carbon remains neutral and halogen remains neutral

","correct":false,"explanation":"The bond is polar because the halogen attracts the shared electrons more strongly than carbon; electron density does not remain evenly distributed. Carbon is therefore δ+."},{"text":"

Halogen is more electropositive than carbon
Electron density is pushed toward the carbon
Carbon becomes $\\delta-$ (electron-rich) and halogen becomes $\\delta+$

","correct":false,"explanation":"The halogen attracts bonding electrons rather than donating them to carbon. Consequently, carbon is electron-deficient and δ+, while the halogen is electron-rich and δ-."}]},{"id":"84868553-f6b9-4373-bc30-5a7b0fd06bd6","cardType":"basic","points":1,"front":"What name is given to the reaction where a bond is broken by its reaction with water or aqueous hydroxide?","back":"Hydrolysis.","mcqOptions":[{"text":"Hydrolysis.","correct":true,"explanation":"Hydrolysis is the breaking of a bond by reaction with water or aqueous hydroxide. In haloalkanes, this replaces the halogen with a hydroxyl group."},{"text":"Oxidation.","correct":false,"explanation":"Oxidation involves electron loss or an increase in oxidation state, not specifically bond breaking by water or aqueous hydroxide. That reaction is called hydrolysis."},{"text":"Neutralisation.","correct":false,"explanation":"Neutralisation is a reaction between an acid and a base forming salt and usually water. Bond breaking by water or aqueous hydroxide is hydrolysis."},{"text":"Reduction.","correct":false,"explanation":"Reduction involves electron gain or a decrease in oxidation state. It is not the specific term for bond cleavage caused by water or aqueous hydroxide."}]},{"id":"53ec4b6a-cf64-471d-94fd-d097e60f791e","cardType":"basic","points":2,"front":"In the nucleophilic substitution of a primary haloalkane by $OH^-$, where does the first curly arrow start and end?","back":"Starts at a lone pair on the $O$ of $OH^-$; ends at the $\\delta+$ carbon atom.","mcqOptions":[{"text":"Starts at a lone pair on the $O$ of $OH^-$; ends at the $\\delta+$ carbon atom.","correct":true,"explanation":"The hydroxide ion attacks the electron-deficient carbon. The first curly arrow therefore starts at an oxygen lone pair and ends at the δ+ carbon atom."},{"text":"Starts at the $\\delta+$ carbon atom; ends at a lone pair on the $O$ of $OH^-$.","correct":false,"explanation":"A curly arrow must start from an electron source, such as an oxygen lone pair, not from the positive carbon. It ends at the δ+ carbon atom."},{"text":"Starts at a lone pair on the $O$ of $OH^-$; ends at the halogen atom.","correct":false,"explanation":"The hydroxide oxygen attacks carbon, not the halogen. The first arrow starts at an oxygen lone pair and ends at the δ+ carbon atom."},{"text":"Starts at the $C-X$ bond; ends at the $\\delta+$ carbon atom.","correct":false,"explanation":"The first arrow represents hydroxide attacking carbon, so it cannot begin at the carbon-halogen bond. It starts at an oxygen lone pair and ends at δ+ carbon."}]},{"id":"e73aeaf9-3fb0-4118-8e67-6ac370fd2de5","cardType":"basic","points":2,"front":"During nucleophilic substitution, what happens to the carbon-halogen bond?","back":"It breaks heterolytically, with both electrons moving to the halogen atom.","mcqOptions":[{"text":"It breaks heterolytically, with both electrons moving to the halogen atom.","correct":true,"explanation":"The carbon-halogen bond breaks heterolytically, meaning both bonding electrons go to the more electronegative halogen, producing a halide ion."},{"text":"It breaks homolytically, with one electron moving to each atom.","correct":false,"explanation":"Homolytic fission gives one bonding electron to each atom and forms radicals. Haloalkane substitution involves heterolytic fission, with both electrons going to the halogen."},{"text":"It breaks heterolytically, with both electrons moving to the carbon atom.","correct":false,"explanation":"Both electrons do not move to carbon because the halogen is more electronegative and leaves with the bonding pair. The bond therefore breaks heterolytically."},{"text":"It remains intact, with both electrons staying between the two atoms.","correct":false,"explanation":"The carbon-halogen bond does not remain intact during substitution. It undergoes heterolytic fission, with both bonding electrons moving to the halogen atom."}]},{"id":"aedb9e68-95d2-4020-aea3-698a0fb6c1e9","cardType":"basic","points":1,"front":"What is the purpose of adding ethanol to the reaction mixture when measuring haloalkane hydrolysis rates?","back":"It acts as a co-solvent to allow the water and haloalkane to mix into a single phase.","mcqOptions":[{"text":"It acts as a co-solvent to allow the water and haloalkane to mix into a single phase.","correct":true,"explanation":"Ethanol acts as a co-solvent because haloalkanes and water do not mix well. It allows them to form one phase, making the hydrolysis rate easier to measure."},{"text":"It acts as a catalyst to allow the water and haloalkane to react without mixing into a single phase.","correct":false,"explanation":"Ethanol is not added primarily as a catalyst; it improves mixing between the aqueous and organic reactants. It acts as a co-solvent to create one phase."},{"text":"It acts as a dehydrating agent to remove the water from the haloalkane phase.","correct":false,"explanation":"Ethanol does not remove water from the mixture. Its purpose is to act as a co-solvent, allowing water and haloalkane to mix into a single phase."},{"text":"It acts as a nucleophile to replace the halogen in the haloalkane and form a single phase.","correct":false,"explanation":"Ethanol is not the nucleophile being measured in this investigation. It is used as a co-solvent so water and haloalkane form a single phase."}]},{"id":"1aa494aa-e8e7-4024-af02-887387e0d8e6","cardType":"cloze","points":null,"front":"In the experimental investigation of hydrolysis rates, the actual nucleophile is $\\text{[...]}$, not silver nitrate.","back":"In the experimental investigation of hydrolysis rates, the actual nucleophile is $\\boxed{H_2O}$, not silver nitrate.","mcqOptions":[{"text":"$$H_2O$","correct":true,"explanation":"Water is the nucleophile because its oxygen lone pairs attack the electron-deficient carbon in the haloalkane. Silver nitrate is used to detect released halide ions."},{"text":"$$OH^-$","correct":false,"explanation":"Hydroxide can act as a nucleophile in aqueous hydroxide reactions, but in this experimental investigation the reacting nucleophile is water, while silver nitrate detects halide ions."},{"text":"$$Ag^+$","correct":false,"explanation":"Silver ions from silver nitrate help indicate halide release by forming a precipitate; they do not attack the haloalkane. Water is the actual nucleophile."},{"text":"$$NO_3^-$","correct":false,"explanation":"Nitrate ions are spectator ions from silver nitrate and do not attack the haloalkane. Water supplies the oxygen lone pair used in nucleophilic substitution."}]},{"id":"a28b7c57-a365-463c-90eb-e4fa37aced4b","cardType":"basic","points":2,"front":"Identify the colours of the silver halide precipitates for $Cl^-$, $Br^-$, and $I^-$.","back":"

$AgCl$: White
$AgBr$: Cream
$AgI$: Yellow

","mcqOptions":[{"text":"

$AgCl$: White
$AgBr$: Cream
$AgI$: Yellow

","correct":true,"explanation":"These are the characteristic silver halide precipitate colours: silver chloride is white, silver bromide is cream, and silver iodide is yellow."},{"text":"

$AgCl$: Cream
$AgBr$: White
$AgI$: Yellow

","correct":false,"explanation":"The first two colours have been swapped, a common confusion between silver chloride and silver bromide. The correct sequence is $AgCl$ white, $AgBr$ cream, and $AgI$ yellow."},{"text":"

$AgCl$: Yellow
$AgBr$: Cream
$AgI$: White

","correct":false,"explanation":"This reverses the colours of the chloride and iodide precipitates. The correct sequence is $AgCl$ white, $AgBr$ cream, and $AgI$ yellow."},{"text":"

$AgCl$: White
$AgBr$: Yellow
$AgI$: Cream

","correct":false,"explanation":"The chloride colour is correct, but cream and yellow have been assigned to the wrong halides. The correct sequence is $AgCl$ white, $AgBr$ cream, and $AgI$ yellow."}]},{"id":"71829167-79d8-45f1-b2df-e61bb8c56701","cardType":"basic","points":1,"front":"What is the trend in rate of hydrolysis for haloalkanes from fluorine to iodine?","back":"Rate increases down the group ($I$ is fastest, $F$ is slowest).","mcqOptions":[{"text":"Rate increases down the group ($I$ is fastest, $F$ is slowest).","correct":true,"explanation":"Hydrolysis becomes faster down the group because the carbon-halogen bond becomes weaker. Iodoalkanes react fastest, while fluoroalkanes react slowest."},{"text":"Rate decreases down the group ($F$ is fastest, $I$ is slowest).","correct":false,"explanation":"This reverses the trend and focuses incorrectly on fluorine's high electronegativity. Hydrolysis rate increases down the group because the carbon-halogen bond weakens, making $I$ fastest and $F$ slowest."},{"text":"Rate increases down the group ($F$ is fastest, $I$ is slowest).","correct":false,"explanation":"The stated direction conflicts with the parenthetical order. Rate increases down the group, so $I$ is fastest and $F$ is slowest because the carbon-halogen bond becomes weaker."},{"text":"Rate decreases down the group ($I$ is fastest, $F$ is slowest).","correct":false,"explanation":"Although the fastest and slowest haloalkanes are identified correctly, the trend is labelled backwards. The rate increases down the group as the carbon-halogen bond weakens."}]},{"id":"57e63f5a-45d8-4708-8b5b-dab7c487fc15","cardType":"basic","points":3,"front":"Why do iodoalkanes react faster than chloroalkanes?","back":"

$C-I$ has a lower bond enthalpy than $C-Cl$
The bond breaks more easily
Bond enthalpy outweighs the effect of bond polarity

","mcqOptions":[{"text":"

$C-I$ has a lower bond enthalpy than $C-Cl$
The bond breaks more easily
Bond enthalpy outweighs the effect of bond polarity

","correct":true,"explanation":"The larger iodine atom forms a longer, weaker $C-I$ bond than the $C-Cl$ bond. Its lower bond enthalpy makes bond breaking easier, outweighing the effect of bond polarity."},{"text":"

$C-I$ has a higher bond enthalpy than $C-Cl$
The bond breaks more easily
Bond polarity outweighs the effect of bond enthalpy

","correct":false,"explanation":"A higher bond enthalpy would make bond breaking harder, and polarity is not the dominant factor here. The correct explanation is that $C-I$ has lower bond enthalpy and breaks more easily."},{"text":"

$C-I$ has a lower bond enthalpy than $C-Cl$
The bond breaks more easily
Bond polarity outweighs the effect of bond enthalpy

","correct":false,"explanation":"The bond strength points are correct, but the final comparison is reversed. Bond enthalpy outweighs the effect of bond polarity, explaining why iodoalkanes hydrolyse faster."},{"text":"

$C-I$ has a higher bond enthalpy than $C-Cl$
The bond breaks more easily
Bond enthalpy outweighs the effect of bond polarity

","correct":false,"explanation":"The tempting error is treating the larger iodine atom as forming a stronger bond. In fact, $C-I$ has lower bond enthalpy than $C-Cl$, so it breaks more easily."}]},{"id":"d3952a83-333b-446d-bba9-e6ae45730d3e","cardType":"cloze","points":null,"front":"When $CFCs$ reach the stratosphere, [ ] provides the energy to break the $C-Cl$ bond.","back":"When $CFCs$ reach the stratosphere, ultraviolet (UV) radiation provides the energy to break the $C-Cl$ bond.","mcqOptions":[{"text":"ultraviolet (UV) radiation","correct":true,"explanation":"Ultraviolet radiation from the Sun supplies enough energy to break the $C-Cl$ bond in CFC molecules when they reach the stratosphere."},{"text":"infrared (IR) radiation","correct":false,"explanation":"Infrared radiation is mainly associated with molecular vibrations and does not provide the relevant photodissociation energy here. Ultraviolet (UV) radiation breaks the $C-Cl$ bond."},{"text":"visible radiation","correct":false,"explanation":"Visible light has lower photon energy than the ultraviolet radiation needed to break the $C-Cl$ bond. Ultraviolet (UV) radiation supplies the necessary energy in the stratosphere."},{"text":"microwave radiation","correct":false,"explanation":"Microwave radiation causes rotational excitation and does not have sufficient energy for this bond-breaking photodissociation. Ultraviolet (UV) radiation provides the required energy."}]},{"id":"d8c9c233-7640-46e0-8789-1bb956bb5ab5","cardType":"cloze","points":null,"front":"When $CFCs$ reach the stratosphere, ultraviolet (UV) radiation provides the energy to break the $\\text{[...]}$ bond.","back":"When $CFCs$ reach the stratosphere, ultraviolet (UV) radiation provides the energy to break the $\\boxed{C-Cl}$ bond.","mcqOptions":[{"text":"$$C-Cl$","correct":true,"explanation":"Stratospheric ultraviolet radiation breaks the carbon-chlorine bond in a CFC, producing a chlorine radical that can begin the catalytic destruction of ozone."},{"text":"$$C-F$","correct":false,"explanation":"The $C-F$ bond is too strong to be broken by stratospheric UV light. The bond broken in this process is $C-Cl$, releasing a chlorine radical."},{"text":"$$C-C$","correct":false,"explanation":"The carbon-carbon bond is not the bond responsible for releasing chlorine radicals from CFCs. Ultraviolet radiation breaks the weaker $C-Cl$ bond instead."},{"text":"$$C-H$","correct":false,"explanation":"CFCs involved in this reaction contain chlorine-carbon bonds rather than the relevant $C-H$ bond. UV radiation breaks $C-Cl$ to form a chlorine radical."}]},{"id":"79971a31-3464-4d87-a73d-45e2ba20cab0","cardType":"basic","points":1,"front":"Why is the $C-F$ bond in a $CFC$ not broken by stratospheric UV light?","back":"The $C-F$ bond enthalpy is too strong ($467 \\text{ kJ mol}^{-1}$).","mcqOptions":[{"text":"The $C-F$ bond enthalpy is too strong ($467 \\text{ kJ mol}^{-1}$).","correct":true,"explanation":"The $C-F$ bond has a high bond enthalpy of $467 \\text{ kJ mol}^{-1}$, so stratospheric UV light does not provide enough energy to break it."},{"text":"The $C-F$ bond enthalpy is too weak ($467 \\text{ kJ mol}^{-1}$).","correct":false,"explanation":"The numerical value is high, not weak, so the description contradicts the chemistry. The $C-F$ bond enthalpy is too strong for stratospheric UV light to break it."},{"text":"The $C-F$ bond enthalpy is too strong ($347 \\text{ kJ mol}^{-1}$).","correct":false,"explanation":"The bond-strength reasoning is correct, but the quoted value is wrong. The $C-F$ bond enthalpy is $467 \\text{ kJ mol}^{-1}$, which is too high for stratospheric UV light to break."},{"text":"The $C-F$ bond enthalpy is too weak ($347 \\text{ kJ mol}^{-1}$).","correct":false,"explanation":"Both the bond description and the numerical value are incorrect. The $C-F$ bond enthalpy is $467 \\text{ kJ mol}^{-1}$ and is too strong to be broken by stratospheric UV light."}]},{"id":"e7861160-9de5-4e86-a281-e3a0d1e8b12c","cardType":"cloze","points":null,"front":"The formation of a chlorine radical from $CF_2Cl_2$ is an example of [ ] or homolytic fission.","back":"The formation of a chlorine radical from $CF_2Cl_2$ is an example of photodissociation or homolytic fission.","mcqOptions":[{"text":"photodissociation","correct":true,"explanation":"Photodissociation is the breaking of a bond using light energy. In this case, UV radiation causes homolytic fission of the CFC's $C-Cl$ bond to form a chlorine radical."},{"text":"photoionisation","correct":false,"explanation":"Photoionisation involves removing an electron to form an ion, whereas this process splits the $C-Cl$ bond into radicals. The correct term is photodissociation, specifically homolytic fission."},{"text":"hydrolysis","correct":false,"explanation":"Hydrolysis is bond breaking by reaction with water and is not driven by UV radiation in the stratosphere. The formation of the chlorine radical is photodissociation by homolytic fission."},{"text":"electrolysis","correct":false,"explanation":"Electrolysis uses electrical energy to drive chemical change, not light energy to split a covalent bond. UV-induced bond breaking in a CFC is called photodissociation."}]},{"id":"d27635aa-ddca-4437-9ade-8593d0052bc8","cardType":"basic","points":1,"front":"State the first propagation step in the catalytic destruction of ozone by $Cl^\\bullet$.","back":"$$Cl^\\bullet + O_3 \\to ClO^\\bullet + O_2$","mcqOptions":[{"text":"$$Cl^\\bullet + O_3 \\to ClO^\\bullet + O_2$","correct":true,"explanation":"A chlorine radical reacts with ozone to form a chlorine monoxide radical and oxygen. This is the first propagation step in the catalytic ozone-destruction cycle."},{"text":"$$Cl^\\bullet + O_2 \\to ClO^\\bullet + O$","correct":false,"explanation":"This incorrectly uses oxygen rather than ozone as the reactant. The first propagation step is $Cl^\\bullet + O_3 \\to ClO^\\bullet + O_2$, where ozone is consumed."},{"text":"$$ClO^\\bullet + O_2 \\to Cl^\\bullet + O_3$","correct":false,"explanation":"This is the reverse direction of the relevant reaction and does not show chlorine radicals attacking ozone. The first propagation step consumes $Cl^\\bullet$ and $O_3$ to form $ClO^\\bullet$ and $O_2$."},{"text":"$$Cl^\\bullet + O_3 \\to ClO^\\bullet + O$","correct":false,"explanation":"The reactants are correct, but ozone contains three oxygen atoms, so the products must conserve them as $ClO^\\bullet + O_2$. Producing only atomic oxygen is unbalanced."}]},{"id":"25e91463-c293-45c6-9784-503424dc4491","cardType":"basic","points":1,"front":"State the second propagation step that regenerates the chlorine radical catalyst.","back":"$$ClO^\\bullet + O \\to Cl^\\bullet + O_2$","mcqOptions":[{"text":"$$ClO^\\bullet + O \\to Cl^\\bullet + O_2$","correct":true,"explanation":"The chlorine monoxide radical reacts with an oxygen atom to form oxygen gas, regenerating $Cl^\\bullet$. Regeneration allows the chlorine radical to catalyse repeated ozone-destruction cycles."},{"text":"$$Cl^\\bullet + O_3 \\to ClO^\\bullet + O_2$","correct":false,"explanation":"This is the first propagation step, in which a chlorine radical reacts with ozone. The second step is $ClO^\\bullet + O \\to Cl^\\bullet + O_2$, which regenerates the catalyst."},{"text":"$$ClO^\\bullet + O_3 \\to Cl^\\bullet + 2O_2$","correct":false,"explanation":"Reacting with ozone rather than an oxygen atom gives the wrong second propagation equation. Catalyst regeneration occurs through $ClO^\\bullet + O \\to Cl^\\bullet + O_2$."},{"text":"$$ClO^\\bullet + O \\to ClO_2^\\bullet$","correct":false,"explanation":"Forming chlorine dioxide would consume the chlorine monoxide radical without regenerating $Cl^\\bullet$. The required step is $ClO^\\bullet + O \\to Cl^\\bullet + O_2$."}]},{"id":"2eac6e3e-dbfc-408f-84c8-e592f5aff897","cardType":"cloze","points":null,"front":"The overall equation for the destruction of ozone in the stratosphere is $\\text{[...]}$.","back":"The overall equation for the destruction of ozone in the stratosphere is $\\boxed{O_3 + O \\to 2O_2}$.","mcqOptions":[{"text":"$$O_3 + O \\to 2O_2$","correct":true,"explanation":"Adding the two propagation steps cancels the regenerated chlorine species, leaving one ozone molecule and one oxygen atom converted into two oxygen molecules."},{"text":"$$2O_2 \\to O_3 + O$","correct":false,"explanation":"This reverses the net ozone-destruction reaction and therefore represents ozone formation. The catalytic cycle has the overall equation $O_3 + O \\to 2O_2$."},{"text":"$$2O_3 \\to 3O_2$","correct":false,"explanation":"This is the overall equation for ozone reacting with another ozone molecule, not the stated chlorine-radical cycle involving atomic oxygen. The required equation is $O_3 + O \\to 2O_2$."},{"text":"$$O_3 \\to O_2 + O$","correct":false,"explanation":"This incorrectly treats atomic oxygen as a product rather than a reactant. Cancelling the catalyst and intermediate gives the net equation $O_3 + O \\to 2O_2$."}]},{"id":"aaa4a0f3-d3a2-41b7-b276-8ee3466be6bb","cardType":"basic","points":2,"front":"Besides $Cl^\\bullet$, identify another radical catalyst that destroys the ozone layer and its source.","back":"

Nitrogen monoxide radical ($^\\bullet NO$)
Source: Lightning strikes or aircraft engines

","mcqOptions":[{"text":"

Nitrogen monoxide radical ($^\\bullet NO$)
Source: Lightning strikes or aircraft engines

","correct":true,"explanation":"Nitrogen monoxide acts as a radical catalyst and is regenerated during ozone destruction. It can enter the stratosphere following formation by lightning or in high-temperature aircraft engines."},{"text":"

Nitrogen dioxide radical ($^\\bullet NO_2$)
Source: Lightning strikes or aircraft engines

","correct":false,"explanation":"Nitrogen dioxide is an intermediate formed from nitrogen monoxide, not the regenerated catalyst identified here. The catalyst is $^\\bullet NO$, sourced from lightning strikes or aircraft engines."},{"text":"

Hydroxyl radical ($^\\bullet OH$)
Source: Chlorofluorocarbons in the stratosphere

","correct":false,"explanation":"Chlorofluorocarbon photodissociation releases chlorine radicals, not hydroxyl radicals. The other catalyst required is nitrogen monoxide, produced by lightning strikes or high-temperature aircraft engines."},{"text":"

Chlorine monoxide radical ($ClO^\\bullet$)
Source: Lightning strikes or aircraft engines

","correct":false,"explanation":"Chlorine monoxide is an intermediate in the chlorine cycle, and lightning or aircraft engines instead produce nitrogen oxides. The alternative catalyst is $^\\bullet NO$."}]},{"id":"e922ad69-2c9a-4a30-8b67-3feb4ac948bb","cardType":"basic","points":1,"front":"State the first propagation step for the destruction of ozone by $^\\bullet NO$.","back":"$$^\\bullet NO + O_3 \\to {^\\bullet NO_2} + O_2$","mcqOptions":[{"text":"$$^\\bullet NO + O_3 \\to {^\\bullet NO_2} + O_2$","correct":true,"explanation":"The nitrogen monoxide radical reacts with ozone, transferring an oxygen atom to form nitrogen dioxide while the remaining oxygen atoms form $O_2$."},{"text":"$$^\\bullet NO_2 + O \\to {^\\bullet NO} + O_2$","correct":false,"explanation":"This is the second propagation step because it regenerates $^\\bullet NO$ from nitrogen dioxide. The first step is $^\\bullet NO + O_3 \\to {^\\bullet NO_2} + O_2$."},{"text":"$$^\\bullet NO + O_2 \\to {^\\bullet NO_2} + O$","correct":false,"explanation":"Using molecular oxygen instead of ozone misses the ozone-destruction step. The first propagation reaction is $^\\bullet NO + O_3 \\to {^\\bullet NO_2} + O_2$."},{"text":"$$^\\bullet NO + O \\to {^\\bullet NO_2}$","correct":false,"explanation":"Atomic oxygen is used during the second propagation step, not the first. Initially, $^\\bullet NO$ reacts with ozone to produce $^\\bullet NO_2$ and $O_2$."}]},{"id":"b113d8cd-77e3-4d74-850b-eadd2a9eff85","cardType":"cloze","points":null,"front":"A single chlorine radical can destroy up to $\\text{[...]}$ ozone molecules because the catalyst is regenerated in the cycle.","back":"A single chlorine radical can destroy up to $\\boxed{100,000}$ ozone molecules because the catalyst is regenerated in the cycle.","mcqOptions":[{"text":"$$100,000$","correct":true,"explanation":"Because the chlorine radical is regenerated rather than consumed, it can repeat the catalytic cycle many times and destroy up to approximately 100,000 ozone molecules."},{"text":"$$10,000$","correct":false,"explanation":"This underestimates the quoted maximum by a factor of ten, commonly from dropping one zero. A regenerated chlorine radical can destroy up to $100,000$ ozone molecules."},{"text":"$$1,000$","correct":false,"explanation":"This underestimates the quoted maximum by a factor of one hundred and fails to reflect the radical’s extensive reuse. The accepted figure is up to $100,000$ molecules."},{"text":"$$1,000,000$","correct":false,"explanation":"This overestimates the quoted maximum by a factor of ten, typically through adding an extra zero. The accepted figure is up to $100,000$ ozone molecules."}]},{"id":"8fcf1ede-16c4-44c5-a139-aa21000ec851","cardType":"cloze","points":null,"front":"A single chlorine radical can destroy up to $100,000$ ozone molecules because the catalyst is [ ] in the cycle.","back":"A single chlorine radical can destroy up to $100,000$ ozone molecules because the catalyst is regenerated in the cycle.","mcqOptions":[{"text":"regenerated","correct":true,"explanation":"The chlorine radical is re-formed during the second propagation step, so it can begin another cycle instead of being used up after destroying one ozone molecule."},{"text":"consumed","correct":false,"explanation":"A catalyst is not consumed in the overall reaction. The chlorine radical is regenerated during propagation, allowing one radical to destroy many ozone molecules."},{"text":"neutralised","correct":false,"explanation":"Neutralisation describes an acid-base reaction and would prevent further catalytic cycles. The chlorine radical is regenerated, so it remains available to attack more ozone."},{"text":"permanently bonded","correct":false,"explanation":"Permanent bonding would remove the radical from the catalytic cycle rather than sustain it. The chlorine radical is regenerated in the propagation steps and can react again."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Haloalkanes","title":"Haloalkanes Flashcards","topbar":{"backHref":"/a-level/chemistry/ocr/a/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"45c20e17-9977-57ba-8d0e-b91405e3f738","walled":"$L39"}],"$L3a","$L3b"]